Timeline for Combinatorial identities
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2013 at 20:30 | answer | added | Mark Wildon | timeline score: 12 | |
| Nov 27, 2013 at 16:48 | answer | added | Ira Gessel | timeline score: 13 | |
| Nov 27, 2013 at 16:47 | comment | added | Shahrooz | It is just a point of view. Suppose we want to construct binary words with length $4n+1$ that one half of these words has weight $n$ and the total weight of these words are greater or equal than $n$. Also, we need the half of these words with this property. This number can be obtained with the left hand side. For the right hand side, we choose $k$ positions from $4n+1$ positions and then from the last selected position (that is 1), we move $n+1$ positions forward (one way for obtaining a word by one half weight greater than $n$) and then choose $n-k$ positions among $3n-k$ remaining positions. | |
| Nov 27, 2013 at 14:27 | comment | added | Daniel Soltész | Well computational evidence suggests that the formula from my previous comment tends to have ridiculously large prime factors for some $u=k \, , \, v=k+1 \, , \, d=u+v+1$. I got seriously distracted by this. | |
| Nov 27, 2013 at 14:14 | comment | added | Daniel Soltész | By rearranging the left hand side it can be transformed into the left hand side of the following identity without alternation: mathoverflow.net/questions/149574/… (With parameters $u=v=n \, , \, d=2n+1$). But computational evidence suggests that for general $u,v,d$ there is no nice formula for it (it has a ridiculously large prime factor for u=13,v=12,d=26). How did you come up with this identity? | |
| Nov 27, 2013 at 10:14 | answer | added | joro | timeline score: 1 | |
| Nov 27, 2013 at 9:31 | comment | added | Dietrich Burde | See also mathoverflow.net/questions/78870/…. | |
| Nov 27, 2013 at 9:05 | review | First posts | |||
| Nov 27, 2013 at 9:07 | |||||
| Nov 27, 2013 at 8:45 | history | asked | Marbor | CC BY-SA 3.0 |